Answer:
[tex]Q=1.9x10^8J[/tex]
Explanation:
Hello!
In this case, since we first know the heat and mass of the first experiment, we are able to compute enthalpy of vaporization of gasoline as shown below:
[tex]Q=m\Delta _{vap} H\\\\\Delta _{vap} H=\frac{11kcal}{1.0g} =11\frac{kcal}{g}[/tex]
Next, we need to the compute the grams in 1.5 gallons of gasoline via its density with the proper units:
[tex]m_{gasoline}=1.5gal*\frac{3785.41mL}{1gal}*\frac{0.74g}{1mL} =4202g[/tex]
Therefore, the new joules are computed via:
[tex]Q=4202g*11\frac{kcal}{g}*\frac{1000cal}{1kcal}*\frac{4.184J}{1cal}\\\\Q=1.9x10^8J[/tex]
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